【发布时间】：2017-10-27 09:31:54
【问题描述】：
问题出现在find函数中，erlang shell告诉我发生了异常错误，它说：
Exception error: no function clause matching seminar1:find("t", []) (seminar1.erl, line 117) in function seminar1:encode/3 ( seminar1.erl, line 113).
我相信发生的事情是在第一个 find 函数中完成的模式匹配总是失败，尽管我不明白为什么，因为手动进行比较的尝试已经成功。
-module(seminar1).
-compile(export_all).
sample() -> "the quick brown fox jumps over the lazy dog
this is a sample text that we will use when we build
up a table we will only handle lower case letters and
no punctuation symbols the frequency will of course not
represent english but it is probably not that far off".
text() -> "this is something that we should encode".
test() ->
Sample = sample(),
Tree = tree(Sample),
Encode = encode_table(Tree),
Decode = decode_table(Tree),
Text = text(),
Seq = encode(Text, Encode),
Text = decode(Seq, Decode).
tree(Sample) -> Freq = freq(Sample),
F = fun({node,N1,V1,_,_}, {node,N2,V2,_,_}) ->
if
V1 > V2 -> false;
V1 == V2 -> if
N1 > N2 -> false;
true -> true
end;
true -> true
end
end,
%lists:sort(F,Freq).
huffman(lists:sort(F,Freq)).
% Calculate the frequency of each letter in the Sample and return a
datastructure of nodes containing the letter involved,
% frequency of it in the sample.
% datastructure {node, Key, Value, Left, Right}
freq(Sample) -> freq(Sample, []).
freq([], Freq) -> Freq;
freq([Char|Rest], Freq) -> freq(Rest, check(Char, Freq)).
% Check function complements the Freq function, it takes the current input
and pattern matches it with the frequency datastructue being built.
% If it scores a hit that particular node has its frequency incremented and
then the whole datastructure is returned.
check(Key, []) -> [{node, [Key], 1, nil, nil}];
check(Key, [{node, [Key], Value, nil, nil}| Tail]) -> [{node, [Key], Value +
1, nil, nil}| Tail];
check(Key, [H|T]) -> [H |check(Key, T)].
% Creates the Huffman tree that is later used to encode a sample.
% The input is the SORTED datastructure derived from the freq-function.
% The leaves of the huffman tree are where actual values reside, branches
are just nodes containing information.
huffman( [ Tree | [] ] ) -> Tree;
huffman([{node, LeftKey, LeftValue, _L1, _R1},
{node, RightKey, RightValue, _L2, _R2} |Tail]) ->
% Creating a branch node
BranchNode = {node, LeftKey ++ RightKey, LeftValue + RightValue, {node,
LeftKey, LeftValue, _L1, _R1}, {node, RightKey, RightValue, _L2, _R2}},
huffman(insert(BranchNode, Tail)).
% A complementary function to the huffman function, inserts the newly made
branchnode into the already sorted tail.
% This is to prevent the sorted tail from becoming unsorted when turning the
tail list into a tree.
% It is inserted as such that the branchnode is the first selection of its
current value,
% meaning that if you have 4 nodes of value 5 ( one being a branchnode) then
the branch node will be the first option.
% This will make the Tree structure left leaning.
%
% N
% N N
% N N
% N N N N
insert(Node, []) -> [Node|[]];
insert(Node, [H|T]) ->
{_, _, Nvalue, _, _} = Node,
{_, _, Hdvalue, _, _} = H,
if
Nvalue =< Hdvalue -> [ Node | [H|T]];
true -> [H | insert(Node, T)]
end.
% Takes the tree created by the huffman-function as input and traverses said
tree.
% Returns a list containing the letters found and their position in the
tree, Left = 0, Right = 1.
% {"e"/[101], [0,0,0]} -- {[Key], [pathway]}
% Traversal method used: Left based traversal.
encode_table(RootNode) -> encode_table(RootNode, [], []).
% When traversing the Tree I need to know the branchnode I am in, the result
list as I am adding letters to it and a PathwayList which is the current
binary path to the branchnode I am in.
encode_table({_, Key, _, nil, nil}, AccList, PathwayList) ->
[AccList | [{Key, reverse(PathwayList)}]];
encode_table({_, _, _, Left, Right}, AccList, PathwayList) ->
encode_table(
Right,
encode_table(Left, AccList, [0| PathwayList]),
[1|PathwayList]).
% Complementary function for the encode_table/3 function, when traversing
the tree the the pathway gets reversed so it needs to be corrected.
reverse(L) -> reverse(L, []).
reverse([], Rev) -> Rev;
reverse([H|T], Rev) -> reverse(T, [H|Rev]).
% Takes a sample text and encodes it in accordance to the encoding table
supplied
encode(Text, Table) -> encode(Text, Table, []).
encode([], _, EncodedText) -> EncodedText;
encode([Letter|Rest], Table, EncodedText) ->
encode(Rest, Table, [find([Letter], Table) | EncodedText]).
% Complementary function to encode/3, searches the Table for the related
Letters binary path.
%find(Letter, []) -> Letter;
find(Letter, [{Letter, BinaryPath} | _Rest]) ->
BinaryPath;
find(Letter, [ _ | Rest]) ->
find(Letter, Rest).
decode_table(tree) -> ok.
decode(sequence, table) -> ok.
test(Letter, [{Letter, Asd} | []]) ->
true;
test(_, _) -> false.
【问题讨论】：
-
要分析的代码很多，但是如果取消注释
%find(Letter, []) -> Letter;这一行，至少错误会消失 -
嗯，是的，这是真的！但是它并不能解决我遇到的问题，即“find(Letter, [{Letter, BinaryPath} | _Rest]) -> BinaryPath;”的模式匹配。一直失败导致程序只遍历表直到最后。